A165617 a(n) is the number of positive integers k such that k is equal to the number of 1's in the digits of the base-n expansion of all positive integers <= k.
2, 4, 8, 4, 21, 5, 45, 49, 83, 10, 269, 11, 202, 412, 479, 15, 1108, 15, 1545, 1219, 1343, 21, 8944, 706, 1043, 5077, 4084, 28, 27589, 27, 32160, 10423, 6689
Offset: 2
Examples
a(3)=4 since there are four values of k such that k is equal to the number of 1's in the digits of the base-3 expansion of all numbers <= k, namely, 1, 4, 5, 12.
From _Jon E. Schoenfield_, Apr 23 2023: (Start)
In the table below, an asterisk appears on each row k at which the cumulative count of 1's in the base-3 expansion of the positive integers 1..k is equal to k:
.
k #1's cume
---------- ---- ----
1 = 1_3 1 1*
2 = 2_3 0 1
3 = 10_3 1 2
4 = 11_3 2 4*
5 = 12_3 1 5*
6 = 20_3 0 5
7 = 21_3 1 6
8 = 22_3 0 6
9 = 100_3 1 7
10 = 101_3 2 9
11 = 102_3 1 10
12 = 110_3 2 12*
(End)
Links
- Tanya Khovanova and Gregory Marton, Archive Labeling Sequences, arXiv:2305.10357 [math.HO], 2023. See p. 9.
Programs
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Mathematica
nn = 7; Table[c = q = 0; Do[c += DigitCount[i, n, 1]; If[c == i, q++], {i, (#^# - #)/(# - 1) &[n]}]; q, {n, 2, nn}] (* Michael De Vlieger, May 24 2023 *) -
PARI
a(n) = {my(nmax = (n^n - 1)/(n - 1) - 1, s = 0, nb = 0); for (i=1, nmax, my(digs = digits(i, n)); s += sum (k=1, #digs, (digs[k] == 1)); if (s == i, nb++);); nb;} \\ Michel Marcus, Aug 20 2013; corrected Apr 23 2023
Extensions
Example corrected by Martin J. Erickson (erickson(AT)truman.edu), Sep 25 2009
Definition and a(10) corrected by Tanya Khovanova, Apr 23 2023
a(11)-a(35) from Gregory Marton, Jul 29 2023
Comments